SOME ESTIMATIONS FOR THE TAYLOR’S REMAINDER‡

In this paper, we establish several integral inequalities for the Taylor’s remainder by Grüss and Cheyshev inequalities. MSC 2000. 26D15.


INTRODUCTION
For two given integrable functions f and g on [a, b], the Chebychev functional T (f, g) is defined by In 1935, Grüss [1] proved that where M, m, L and l are constants.Inequality (1.1) is called Grüss inequality.The well-known Chebyshev inequality [2] can be stated as follows: if both f and g are increasing or decreasing, then T (f, g) ≥ 0.
If one of the functions f and g is increasing and the other decreasing, then the above inequality is reversed.
In what follows n denotes a non-negative integer.We denote by R n,f (x 0 , x) the nth Taylor remainder of the function f (x) with center x 0 , i.e.
The Taylor remainder has been the subject of intensive research [3]- [9].In particular, many remarkable integral inequalities for the Taylor remainder can be found in the literature [5]- [7].The following Theorems A and B were proved by Gauchman in [6]. and It is the aim of this paper to establish several new inequalities for the Taylor remainder by Grüss and Cheyshev inequalities.

MAIN RESULTS
Lemma 2.1.Let f (x) be a function defined on [a, b] and Proof.We only give the proof of (2.1) in detail, the similar argument leads to (2.2).It follows from the formula of integration by parts that b x 0 Then for any x ∈ [x 0 , b], we clearly see that 2) and inequality (2.6) imply that . Therefore, inequality (2.3) follows form inequalities (2.5) and (2.7).
If take n = 1 in Theorem 2.2, then we have In particular, if x 0 = a+b 2 , then Corollary 2.3 becomes where m 1 , m 2 , M 1 and M 2 are constants.Then In particular, if x 0 = a+b 2 , then Corollary 2.5 becomes Corollary 2.6.
where m 1 , m 2 , M 1 and M 2 are constants.Then (1) If n is an odd number and f (n+1) (x) is increasing in [a, b], then (3) If n is an even number and Proof.We divide the proof into two cases.
n+1)! , then we clearly see that G(x) is decreasing in [x 0 , b].We divide this case into two subcases.
From the monotonicity of F and G we have Therefore, inequalities (2.5) and (2.12) lead to the conclusion that Then equation (2.1) and inequality (2.5) lead to the conclusion that (n+1)! .We divide the discussion into four subcases.